Exercise 1527Combinatorics › Binomial coefficientsL1Added 3 Nov 2026 An identity with inverses #inverse binomial#identity#integral StatementShow that ∑k=0n(−1)kk+1(nk)=1n+1\sum_{k=0}^{n}\frac{(-1)^k}{k+1}\binom{n}{k}=\frac{1}{n+1}∑k=0nk+1(−1)k(kn)=n+11. HintUse 1k+1(nk)=1n+1(n+1k+1)\frac{1}{k+1}\binom{n}{k}=\frac{1}{n+1}\binom{n+1}{k+1}k+11(kn)=n+11(k+1n+1). Show solutionHide solutionSolutionIdentity. 1k+1(nk)=1n+1(n+1k+1)\frac{1}{k+1}\binom{n}{k}=\frac{1}{n+1}\binom{n+1}{k+1}k+11(kn)=n+11(k+1n+1).Sum.∑k=0n(−1)kk+1(nk)=1n+1∑k=0n(−1)k(n+1k+1).\sum_{k=0}^{n}\frac{(-1)^k}{k+1}\binom{n}{k}=\frac{1}{n+1}\sum_{k=0}^{n}(-1)^k\binom{n+1}{k+1}.k=0∑nk+1(−1)k(kn)=n+11k=0∑n(−1)k(k+1n+1).Alternating sum. Let j=k+1j=k+1j=k+1: ∑j=1n+1(−1)j−1(n+1j)=−(1−1)n+1+1=1\sum_{j=1}^{n+1}(-1)^{j-1}\binom{n+1}{j}=-(1-1)^{n+1}+1=1∑j=1n+1(−1)j−1(jn+1)=−(1−1)n+1+1=1 (using ∑j=0n+1(−1)j(n+1j)=0\sum_{j=0}^{n+1}(-1)^j\binom{n+1}{j}=0∑j=0n+1(−1)j(jn+1)=0, so ∑j=1n+1(−1)j−1(n+1j)=1\sum_{j=1}^{n+1}(-1)^{j-1}\binom{n+1}{j}=1∑j=1n+1(−1)j−1(jn+1)=1).Conclusion. Total =1n+1=\frac{1}{n+1}=n+11.